[tex]\left\{\begin{matrix} \sqrt{x^2-(x+y)}=\frac{y}{\sqrt[3]{x-y}} & \\ 2(x^2+y^2)-3\sqrt{2x-1}=11& \end{matrix}\right.[/tex]
[tex]ĐK : x\geq \frac{1}{2}[/tex]
[tex]pt(1)\iff \sqrt{x^2-x-y}-y+y\frac{\sqrt[3]{x-y}-1}{\sqrt[3]{x-y}}=0 \iff (x-y-1)((\frac{x+y}{\sqrt{x^2-x-y}+y})+\frac{y}{\sqrt[3]{x-y}((\sqrt[3]{x-y})^2+\sqrt[3]{x-y}+1)})=0[/tex]
Xét $y<0$ ta có :
$pt(1)\Rightarrow \frac{y}{\sqrt[3]{x-y}}>0$
$y<0 \Leftrightarrow x<y<0$ mà $x\geq \frac{1}{2}$
$\Rightarrow$ vô lí.
Xét y>0 thì cụm bên trong >0 nên $x-1=y$
$pt(2)\iff 2(x^2+(x-1)^2)-3\sqrt{2x-1}-11=0$
$\iff (4x^2-4x-15)-3(\sqrt{2x+1}-2)=0$
$\iff (2x-5)(2x+3)-3\frac{2x-5}{(\sqrt{2x+1}+2)}=0$
$\iff (2x-5)(2x+3-3\frac{1}{(\sqrt{2x+1}+2)})=0$
$\iff(2x-5)(2x+3(\frac{\sqrt{2x+1}+1}{\sqrt{2x+1}+2}))=0$
$\iff x=\frac{5}{2}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
